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I'm reading Qing Liu. I feel confused about the proof of theorem 4.3.12 and corollary 3.14.

Theorem 4.3.12. Let $f$ be a morphism of locally Noetherian schemes. Let $x\in X$ and $y= f(x)$. Then $\dim (\mathcal{O}_{X_y,x})\geq \dim (\mathcal{O}_{X,x})-\dim (\mathcal{O}_{Y,y})$. If, moreover, $f$ is flat, then we have equality. enter image description here

My question:

1.(red sentences) How do we know $t$ is not invertible in $B=\mathcal O_{X,x}$?

2.(green sentence) After base change with respect to $Y'$, $X'\rightarrow X$ is a closed immersion. But why $x$ is in $X' = X\times_YY'$?

Corollary 4.3.14. Let $f:X\rightarrow Y$ be a flat subjective morphism of algebraic varieties. We suppose that $Y$ is irreducible and that $X$ is equidimensional. Then for every $y\in Y$, the fiber $X_y$ is equidimensional, and we have $\dim(X_y) = \dim(X)-\dim(Y)$ enter image description here

My Question: Qing Liu defines algebraic variety as Noetherian scheme of finite type over $k$. That means $X_i$ may not be reduced. But Proposition 2.5.23(a) only works for domains. In this case how to get the highlight equality?

Theorem 2.5.15 Let $(A,\frak m)$ be a Noetherian local ring, $f\in \frak m$. Then we have $\dim(A/fA)\geq \dim(A)-1$. Moreover, equality holds if $f$ is not contained in any minimal prime ideal of $A$.

Theorem 2.5.23(a) Let $A$ be a finitely generated integral domain over $k$. Let $p$ be a prime ideal of $A$, we have $\operatorname{ht}(p) + \dim(A/p) = \dim(A)$.

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  1. $A\to \mathcal{O}_{X,x}$ is a local homomorphism of local rings, so it sends $\mathfrak{m}_A$ to $\mathfrak{m}_x$. $t$ is in the first by assumption, so it's in the second, and the maximal ideal of a local ring is precisely the non-invertible elements.

  2. Since $Y'\to Y$ hits $y$ by construction, there are maps from $\operatorname{Spec} k(x)$ to $Y'$ and $X$ which agree after composing with the map to $Y$. So there's a map $\operatorname{Spec} k(x)\to X'$ which after composing with $X'\to X$ agrees with the map $\operatorname{Spec} k(x)\to X$ which picks out $x$. As $X'\to X$ is a closed immerison, this means $x\in X'$.

  3. Dimension is a purely topological invariant insensitive to the scheme structure. So you can replace your scheme by its reduction.

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  • $\begingroup$ Thank you for your answer. For (2), why there is a morphism $Spec k(x)\rightarrow Y'$(I understand there is $Spec k(y)\rightarrow Y'$). For (3), could we use the same argument to generalize theorem 2.5.23(a) to $A$ not necessary domain? $\endgroup$ – Hydrogen Mar 21 at 0:25
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    $\begingroup$ The fact that $x$ maps to $y$ means there's a map of residue fields $k(y)\to k(x)$, or a map of spectra $\operatorname{Spec} k(x)\to\operatorname{Spec} k(y)$. Compose with the map you know. Secondly, no, not without further adjustments - consider something like $R=k[x,y,z]/(xy,xz)$: this has dimension $2$, but the quotient by the prime ideal $(z)$ gives the ring $k[x,y]/(xy)$ which is of dimension one. Thinking about the geometry of that should be instructive. $\endgroup$ – KReiser Mar 21 at 0:32
  • $\begingroup$ It seems that the statement holds when $Spec(A)$ is irreducible but not necessary reduced. $\endgroup$ – Hydrogen Mar 21 at 3:46
  • $\begingroup$ Sorry. I have one last question. In proof of corollary 3.14, he says 'Since x is a closed point, k(x) is algebraic over k(y)'. I'm confused about this. $\endgroup$ – Hydrogen Mar 21 at 5:19
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    $\begingroup$ 1. Yes, $\operatorname{Spec} A$ irreducible means $A/Nil(A)$ is a domain. 2. $x$ is a closed point in a scheme locally of finite type over a field, apply Zariski's lemma. $\endgroup$ – KReiser Mar 21 at 6:06

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